In an aqueous solution of a certain acid the acid is 0.10% dissociated and the pH is 4.06. Calculate the acid dissociation constant Ka of the acid. Round your answer to 2 significant digits

Answer : The acid dissociation constant Ka of the acid is,

Explanation :

First we have to calculate the concentration of hydrogen ion.

Given: pH = 4.06

The dissociation of acid reaction is:

Initial conc.      c        0         0

At eqm.           c-cα    cα       cα

Given:

Degree of dissociation = α = 0.10 % = 0.001

The expression of dissociation constant of acid is:

Now put all the given values in this expression, we get:

Thus, the acid dissociation constant Ka of the acid is,

1.consider the following balanced equation and answer parts a through. show calculations for each part.  

3 ag(s) + 4 hno3(aq) → 3 agno3(aq) + no(g) + 2 h2o(l)  

a. calculate the number of moles of agno3 produced by the reaction of 3 moles of ag with excess hno3. if the number to the left of the compound represents the number of moles of the compound which are present, then 3 moles of ag react with 4 moles hno3 to produce 3 moles agno3 (aq).

so the answer to a would be 3 moles.

b. calculate the number of moles of agno3 produced by the reaction of 108 grams of ag with excess hno3.  

i'm not sure how to plug in the grams of silver into an equation using moles to find how many moles are produced.

c. calculate the number of grams of agno3 produced by the reaction of 108 grams of ag with excess hno3  

d. calculate the number of moles of no produced as a byproduct of the reaction of 108 grams of ag with excess hno3.  

e. calculate the number of grams of no produced as a byproduct starting with 108 grams of ag and excess hno3.  

explanation: